Derivative of x^x
x^x breaks both the power rule and the exponential rule — the base and the exponent are both variables — so it demands the heaviest tool in the kit: logarithmic differentiation.
Step-by-step solution
Let y = xx. Then ln y = x ln x — the exponent comes down, converting an impossible power into an ordinary product.
Left side (chain rule): y′/y. Right side (product rule): 1·ln x + x·(1/x) = ln x + 1.
y′ = y·(ln x + 1) = xx(ln x + 1).
Why it works
Why do the standard rules fail? The power rule d/dx[xⁿ] = nxⁿ⁻¹ assumes the exponent is frozen; the exponential rule d/dx[aˣ] = aˣ ln a assumes the base is frozen. In x^x nothing is frozen. Amusingly, if you (incorrectly) apply both rules anyway and add the results — x·x^(x−1) + x^x ln x — you get the right answer. That's not luck: it's the multivariable chain rule showing through, with each rule capturing one variable's contribution.
The factor (ln x + 1) also hands you the function's minimum for free: the derivative is zero when ln x = −1, i.e. at x = 1/e ≈ 0.368, where x^x bottoms out at about 0.692. It's a satisfying payoff — a function that looks untouchable yields its minimum to three lines of logarithms.
Domain and a second derivation
The formula on this page uses the smooth real-valued definition xˣ = exp(x ln(x)) for x > 0. Some negative inputs admit individual real powers, but they do not give the same smooth function on an open interval. Do not extend this logarithmic derivative formula to negative inputs.
Differentiate exp(x ln(x)) directly: the outer exponential stays unchanged, and the product rule gives d/dx[x ln(x)] = ln(x) + 1. Their product is xˣ(ln(x) + 1), agreeing with logarithmic differentiation.
At x = 1 the function value and derivative are both 1, so the tangent line is y = x. On 0 < x < 1/e, ln(x) + 1 is negative; on x > 1/e it is positive. Thus x = 1/e is the global minimum on the positive domain, with value exp(−1/e) ≈ 0.6922. The practice formulas for x²ˣ and xˢⁱⁿ⁽ˣ⁾ also assume x > 0.
Common mistakes
- Applying the power rule alone (x·x^(x−1)) or the exponential rule alone (x^x ln x) — each captures only half the change.
- Forgetting the y′/y on the left side when differentiating ln y implicitly.
- Leaving the answer in terms of y instead of substituting x^x back in.
Practice problems
Differentiate x2x
Answer: 2x2x(ln x + 1) — same method; ln y = 2x ln x.
Where is the minimum of xx for x > 0?
Answer: x = 1/e, from setting ln x + 1 = 0.
Differentiate xsin(x)
Answer: xsin(x)[cos(x) ln x + sin(x)/x] — log both sides, product rule.